chemistry 161 chapter 6
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CHEMISTRY 161
Chapter 6
www.chem.hawaii.edu/Bil301/welcome.html
THERMODYNAMICS
HEAT CHANGE
quantitative study of heat and energy changes of a system
the state (condition) of a system is defined by
T, p, n, V, E
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
the state (condition) of a system is defined by
T, p, n, V, E
STATE FUNCTIONS
properties which depend only on the initial and final state, but not on the way how this
condition was achieved
ΔV = Vfinal – Vinitial
Δp = pfinal – pinitial
ΔT = Tfinal – Tinitial
ΔE = Efinal – Einitial
Energy is a STATE FUNCTION
IT DOES NOT MATTER WHICH PATH YOU TAKE
ΔE = m g Δh
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
EN
TH
AL
PY
, H
Reactants
Products
CO2(g) + 2H2O(g)
- 802 kJ
- 88 kJ
- 890 kJ
Hess Law
heat is spontaneous transfer of thermal energy two bodies at different temperatures T1 > T2
spontaneous
T1
T2
Zeroth Law of Thermodynamics
a system at thermodynamical equilibrium
has a constant temperature
First Law of Thermodynamics
energy can be converted from one form to another,
but cannot be created or destroyed
CONSERVATION OF ENERGY
SYSTEM
SURROUNDINGS
THE TOTAL ENERGY OF THE UNIVERSE IS CONSTANT
ΔE = ΔEsystem + ΔEsurrounding = 0
-+
ΔΔEEsystemsystem = = ΔΔQ + Q + ΔΔWWΔΔEEsystemsystem = = ΔΔQ + Q + ΔΔWW
First Law of Thermodynamics
ΔΔQ heat changeQ heat changeΔΔQ heat changeQ heat change ΔΔW work doneW work doneΔΔW work doneW work done
Q > 0 ENDOTHERMICQ > 0 ENDOTHERMIC
Q < 0 EXOTHERMICQ < 0 EXOTHERMIC?
ΔΔW = - p W = - p ΔΔVVΔΔW = - p W = - p ΔΔVV
mechanical workmechanical workmechanical workmechanical work
M
the energy of gas goes up
M
the energy of gas goes down
ΔV < 0ΔV > 0
First Law and Enthalpy
ΔΔEEsystemsystem = = ΔΔQ - pQ - pΔΔVVΔΔEEsystemsystem = = ΔΔQ - pQ - pΔΔVV
1.constant pressure → enthalpy change Δ H
2. ideal gas law → p V = n R T
p ΔV = Δn R T
ΔΔEEsystemsystem = = ΔΔQ – R T Q – R T ΔΔnnΔΔEEsystemsystem = = ΔΔQ – R T Q – R T ΔΔnn
ΔΔQ = Q = ΔΔEEsystemsystem + R T + R T ΔΔn = n = ΔΔHHΔΔQ = Q = ΔΔEEsystemsystem + R T + R T ΔΔn = n = ΔΔHH
ΔΔQ = Q = ΔΔEEsystemsystem + R T + R T ΔΔn = n = ΔΔHHΔΔQ = Q = ΔΔEEsystemsystem + R T + R T ΔΔn = n = ΔΔHH
Δ n = nfinal – ninitial
Calculate the energy change of a system for the reaction process at 1 atm and 25C
2 CO(g) + O2(g) → 2 CO2(g) ΔHo = -566.0 kJ
ΔΔEEsystemsystem = = ΔΔHH00 - R T - R T ΔΔn n ΔΔEEsystemsystem = = ΔΔHH00 - R T - R T ΔΔn n
ΔΔEEsystemsystem = -563.5 kJ = -563.5 kJ ΔΔEEsystemsystem = -563.5 kJ = -563.5 kJ
A gas is compressed in a cylinder from a volume of 20 L to 2.0 L by a constant pressure of 10 atm.
Calculate the amount of work done on the system.
Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 30.0°C.
Zn(s) + 2H+(aq) → Zn2+(aq) + H2 (g)
The heat of solution of KCl is +17.2 kJ/mol and the
lattice energy of KCl(s) is 701.2 kJ/mol. Calculate
the total heat of hydration of 1 mol of gas phase
K+ ions and Cl– ions.
Homework
Chapter 6, problems